Problem 49 - Calculus Single Variable

Question

In each of Exercises 48-53, a function \(f\) is given. In each case, the domain of \(f\) and the value of \(f(x)\) depend on a positive constant \(b\). Determine each point at which \(f\) has a local extremum. Use the First Derivative Test to classify the extremum. $$ f(x)=x^{2} \sqrt{b^{2}-x^{2}}, 0

Step-by-Step Solution

Answer

The local extremum occurs at \( x = \sqrt{\frac{2b^2}{3}} \), and it is a local maximum.

Step 1: Understand the Function

We are given the function \( f(x) = x^2 \sqrt{b^2 - x^2} \) with domain \( 0 < x < b \). We need to find points where \( f \) has a local extremum and use the First Derivative Test to classify them.

Step 2: Find the First Derivative

To find local extrema, we first calculate the derivative \( f'(x) \). Using the product and chain rules, \( f'(x) = \frac{d}{dx}\left(x^2\right)\sqrt{b^2-x^2} + x^2\frac{d}{dx}\left(\sqrt{b^2-x^2}\right)\).Compute:- \( \frac{d}{dx}(x^2) = 2x \)- \( \frac{d}{dx}(\sqrt{b^2-x^2}) = \frac{-x}{\sqrt{b^2-x^2}} \).So, \( f'(x) = 2x\sqrt{b^2-x^2} - \frac{x^3}{\sqrt{b^2-x^2}} \). Combine and simplify to:\[ f'(x) = \frac{2x(b^2 - x^2) - x^3}{\sqrt{b^2-x^2}} = \frac{x(2b^2 - 3x^2)}{\sqrt{b^2-x^2}} \].

Step 3: Set the Derivative to Zero

To find critical points, set \( f'(x) = 0 \):\[ \frac{x(2b^2 - 3x^2)}{\sqrt{b^2-x^2}} = 0 \].This gives us: \( x = 0 \) or \( 2b^2 - 3x^2 = 0 \). Since \( x = 0 \) is not in the domain \( 0 < x < b \), solve \( 2b^2 - 3x^2 = 0 \):\[ x^2 = \frac{2b^2}{3} \], so \( x = \pm\sqrt{\frac{2b^2}{3}} \). Since \( x > 0 \), we choose \( x = \sqrt{\frac{2b^2}{3}} \).

Step 4: Apply the First Derivative Test

We next use the First Derivative Test around the critical point \( x = \sqrt{\frac{2b^2}{3}} \). Examine the sign of \( f'(x) \) around this point.- For \( x < \sqrt{\frac{2b^2}{3}} \), \( f'(x) > 0 \) (verify by substituting into \( f'(x) \))- For \( x > \sqrt{\frac{2b^2}{3}} \), \( f'(x) < 0 \).This indicates that \( x = \sqrt{\frac{2b^2}{3}} \) is a local maximum because \( f'(x) \) changes from positive to negative.

Key Concepts

First Derivative Test

The First Derivative Test is an essential tool in calculus used to identify the nature of critical points for a function. The test helps us uncover where a function might take on maximum or minimum values, known as local extrema.

Here's how it works:

• Calculate the first derivative of your function, which represents the slope of the tangent line at any point.
• Find the critical points by solving the equation where the first derivative equals zero or doesn't exist.
• Choose values slightly less than and greater than each critical point to see how the sign of the derivative changes around these points.
• If the derivative changes from positive to negative, this indicates a local maximum. If it switches from negative to positive, it shows a local minimum.

Applying this test can classify each critical point as a local maximum, minimum, or neither. In the case of the function provided, it helps us identify that the function has a local maximum at the specific critical point.

critical points

Critical points are crucial in identifying where a function might achieve its highest or lowest values locally, called extrema. These are points where the first derivative of the function is either zero or undefined.

To find critical points:

• Take the first derivative of the function.
• Set the derivative equal to zero and solve for the variable.
• Identify points where the derivative does not exist within the domain.

In our exercise, solving the equation \( \frac{x(2b^2 - 3x^2)}{\sqrt{b^2-x^2}} = 0 \) leads us to discover critical points within the domain, which are potential locations of local extrema. Remember, not all critical points result in local maxima or minima; the First Derivative Test confirms which type of extremum, if any, occurs at these points.

domain of a function

Understanding the domain of a function is essential for correctly working with derivatives and finding extrema. The domain specifies all possible input values (x-values) for which the function is defined.

In this context, the domain is limited to \( 0 < x < b \), meaning:

• The function is defined only for values of \( x \) between 0 and \( b \).
• It helps in narrowing down where critical points can lie.
• Ensures that any results from our calculations don't include values outside this range.

This limit plays a vital role in identifying the valid critical points and determining the nature of extrema within those bounds. By understanding the domain, we make sure the calculations lead to meaningful results for the function inside its proper scope.

product rule

The product rule is a differentiation technique used when you need to differentiate the product of two functions. When faced with a function like \( f(x) = x^2 \sqrt{b^2 - x^2} \), the product rule is applied to accommodate the varying parts of the function.

Here's the rule:

• If you have a function \( g(x)h(x) \), the derivative is \( g'(x)h(x) + g(x)h'(x) \).
• This technique helps break down complicated derivatives by treating constituent parts separately.

In this exercise, \( f(x) = x^2 \sqrt{b^2 - x^2} \) is factored into two bits: \( x^2 \) and \( \sqrt{b^2 - x^2} \). Calculating these separately and then combining them using the product rule allows for efficiently determining \( f'(x) \). This step was crucial for analyzing the behavior of the function and finding critical points.